×

Close

- Plasma Physics - PP
- Note
- 5 Topics
**965 Views**- 11 Offline Downloads
- Uploaded 1 year ago

Touch here to read

Page-1

Topic:

Chapter 1 Introduction 1.1 1.1.1 What is a Plasma? An ionized gas A plasma is a gas in which an important fraction of the atoms is ionized, so that the electrons and ions are separately free. When does this ionization occur? When the temperature is hot enough. Balance between collisional ionization and recombination: Figure 1.1: Ionization and Recombination Ionization has a threshold energy. Recombination has not but is much less probable. Threshold is ionization energy (13.6eV, H). χi Integral over Maxwellian distribution gives rate coeﬃcients (reaction rates). Because of the tail of the Maxwellian distribution, the ionization rate extends below T = χi . And in equilibrium, when nions < σi v > = , (1.1) < σr v > nneutrals 6

Figure 1.2: Ionization and radiative recombination rate coeﬃcients for atomic hydrogen the percentage of ions is large (∼ 100%) if electron temperature: Te > ∼ χi /10. e.g. Hydrogen ◦ is ionized for Te > 1eV (11,600 k). At room temp r ionization is negligible. ∼ For dissociation and ionization balance ﬁgure see e.g. Delcroix Plasma Physics Wiley (1965) ﬁgure 1A.5, page 25. 1.1.2 Plasmas are QuasiNeutral If a gas of electrons and ions (singly charged) has unequal numbers, there will be a net charge density, ρ. ρ = ne (−e) + ni (+e) = e(ni − ne ) (1.2) This will give rise to an electric ﬁeld via �.E = ρ e = (ni − ne ) �0 �0 (1.3) Example: Slab. dE ρ = dx �0 x →E = ρ �0 7 (1.4) (1.5)

Figure 1.3: Charged slab This results in a force on the charges tending to expel whichever species is in excess. That is, if ni > ne , the E ﬁeld causes ni to decrease, ne to increase tending to reduce the charge. This restoring force is enormous! Example Consider Te = 1eV , ne = 1019 m−3 (a modest plasma; c.f. density of atmosphere nmolecules ∼ 3 × 1025 m−3 ). Suppose there is a small diﬀerence in ion and electron densities Δn = (ni − ne ) so ρ = Δn e (1.6) Then the force per unit volume at distance x is Fe = ρE = ρ2 x x = (Δn e)2 �0 �0 (1.7) Take Δn/ne = 1% , x = 0.10m. Fe = (1017 × 1.6 × 10−19 )2 0.1/8.8 × 10−12 = 3 × 106 N.m−3 (1.8) Compare with this the pressure force per unit volume ∼ p/x : p ∼ ne Te (+ni Ti ) Fp ∼ 1019 × 1.6 × 10−19 /0.1 = 16N m−3 (1.9) Electrostatic force >> Kinetic Pressure Force. This is one aspect of the fact that, because of being ionized, plasmas exhibit all sorts of col lective behavior, diﬀerent from neutral gases, mediated by the long distance electromagnetic forces E, B. Another example (related) is that of longitudinal waves. In a normal gas, sound waves are propagated via the intermolecular action of collisions. In a plasma, waves can propagate when collisions are negligible because of the coulomb interaction of the particles. 8

1.2 1.2.1 Plasma Shielding Elementary Derivation of the Boltzmann Distribution Basic principle of Statistical Mechanics: Thermal Equilibrium ↔ Most Probable State i.e. State with large number of possible ar rangements of microstates. Figure 1.4: Statistical Systems in Thermal Contact Consider two weakly coupled systems S1 , S2 with energies E1 , E2 . Let g1 , g2 be the number of microscopic states which give rise to these energies, for each system. Then the total number of microstates of the combined system is (assuming states are independent) g = g1 g2 (1.10) If the total energy of combined system is ﬁxed E1 + E2 = Et then this can be written as a function of E1 : and g = g1 (E1 )g2 (Et − E1 ) dg dg1 dg2 g2 − g1 . = dE dE dE1 The most probable state is that for which dg dE1 (1.11) (1.12) = 0 i.e. 1 dg1 1 dg2 d d = or ln g1 = ln g2 g1 dE g2 dE dE dE Thus, in equilibrium, states in thermal contact have equal values of (1.13) d dE ln g. One deﬁnes σ ≡ ln g as the Entropy. d And [ dE ln g]−1 = T the Temperature. Now suppose that we want to know the relative probability of 2 microstates of system 1 in equilibrium. There are, in all, g1 of these states, for each speciﬁc E1 but we want to know how many states of the combined system correspond to a single microstate of S1 . Obviously that is just equal to the number of states of system 2. So, denoting the two values of the energies of S1 for the two microstates we are comparing by EA , EB the ratio of the number of combined system states for S1A and S1B is g2 (Et − EA ) = exp[σ(Et − EA ) − σ(Et − EB )] g2 (Et − EB ) 9 (1.14)

## Leave your Comments